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3 years ago

What is the total mass of products formed when 16 grams of ch4?

1 answer:

3 0

When CH₄ is burnt in excess O₂ following products are formed,

CH₄ + 2 O₂ → CO₂ + 2 H₂O

According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,

1 mole of CO₂ and 2 moles of H₂O

Converting 1 mole CO₂ to grams;
As,
Mass = Moles × M.mass

Mass = 1 mol × 44 g.mol⁻¹

Mass = 40 g of CO₂

Converting 2 moles of H₂O to grams,

Mass = 2 mol × 18 g.mol⁻¹

Mass = 36 g of H₂O

Total grams of products;

Mass of CO₂ = 44 g
+ Mass of H₂O = 36 g
-------------
Total = 80 g of Product

Result:
80 grams of product is formed when 16 grams of CH₄ is burnt in excess of Oxygen.

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

Strike441 [17]

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

5 0

2 years ago

Why do two oxygen atoms form a double covalent bond?

elena-s [515]

A lone oxygen atom has 6 electrons in its outer shell which is not very stable, whereas as full octet (8 outer shell electrons) is stable. In order to achieve this two oxygen atoms will share 4 electrons, each contributing 2 electrons. Since these electrons exist within the orbitals of both atoms, to oxygen atoms essentially achieve a full octet.

5 0

3 years ago

the air pressure from the atmosphere measures 0.5atm at an altitude of 18,000 ft. How much pressure is this in pounds per square

Julli [10]

14.7lb/in²

Explanation:

This is a conversion problem from atm to pounds per square inch

Pressure is the combined force per unit area of substance on another.

Given pressure = 0.5atm

To solve this problem ;

Convert from atm to N/m² to pound per square inch

1 atm = 101325N/m²

0.5atm = 0.5 x 101325 = 50,662.5‬N/m²

now we need to convert to pound per square inch;

1N = 0.225lb

1m = 39.37in

Solving;

101325 N $x \frac{1}{m^{2} } x \frac{0.225lb}{1N } x \frac{m^{2} }{39.37in x 39.37in}$ = 14.7lb/in²

learn more:

Pressure of gases brainly.com/question/10897667

#learnwithBrainly

8 0

3 years ago

What equipment might you use to find the SHC of a material? Need to know the answer ASAP PLZ

algol [13]

Theses can include the power supply circuit a joule meter to measure the energy transferred which makes the calculations a lot easier.

4 0

3 years ago